17.1. THEOREMS BASED ON BAIRE CATEGORY 437

Theorem 17.1.4 Let X and Y be two normed linear spaces and let L : X → Y be linear(L(ax+by) = aL(x)+bL(y) for a,b scalars and x,y ∈ X). The following are equivalent

a.) L is continuous at 0b.) L is continuousc.) There exists K > 0 such that ||Lx||Y ≤ K ||x||X for all x ∈ X (L is bounded).

Proof: a.)⇒b.) Let xn → x. It is necessary to show that Lxn → Lx. But (xn− x)→ 0and so from continuity at 0, it follows

L(xn− x) = Lxn−Lx→ 0

so Lxn→ Lx. This shows a.) implies b.).b.)⇒c.) Since L is continuous, L is continuous at 0. Hence ||Lx||Y < 1 whenever

||x||X ≤ δ for some δ . Therefore, suppressing the subscript on the || ||,

||L(

δx||x||

)|| ≤ 1.

Hence||Lx|| ≤ 1

δ||x||.

c.)⇒a.) follows from the inequality given in c.).

Definition 17.1.5 Let L : X → Y be linear and continuous where X and Y are normedlinear spaces. Denote the set of all such continuous linear maps by L (X ,Y ) and define

||L||= sup{||Lx|| : ||x|| ≤ 1}. (17.1.1)

This is called the operator norm.

Note that from Theorem 17.1.4 ||L|| is well defined because of part c.) of that Theorem.The next lemma follows immediately from the definition of the norm and the assump-

tion that L is linear.

Lemma 17.1.6 With ||L|| defined in 17.1.1, L (X ,Y ) is a normed linear space. Also||Lx|| ≤ ||L|| ||x||.

Proof: Let x ̸= 0 then x/ ||x|| has norm equal to 1 and so∣∣∣∣∣∣∣∣L( x||x||

)∣∣∣∣∣∣∣∣≤ ||L|| .Therefore, multiplying both sides by ||x||, ||Lx|| ≤ ||L|| ||x||. This is obviously a linearspace. It remains to verify the operator norm really is a norm. First of all, if ||L||= 0, thenLx = 0 for all ||x|| ≤ 1. It follows that for any x ̸= 0,0 = L

(x||x||

)and so Lx = 0. Therefore,

L = 0. Also, if c is a scalar,

||cL||= sup||x||≤1

||cL(x)||= |c| sup||x||≤1

||Lx||= |c| ||L|| .